Question

Alice and Bob are scheduled to meet at Quack's coffee shop at a set time, but they are often late for appointments. Each will arrive at Quack's with a delay between 0 and 1 hour, with all pairs of delays being equally likely. Moreover, both Alice and Bob shun the conveniences of the modern world, and thus they do not have cell phones to contact each other. The first person to arrive will wait for 20 minutes (1/3 of an hour) but will get impatient and leave in disgust if the other has not yet arrived by then. What is the probability that they will meet? Show/explain how you arrived at your answer. You can derive the answer any way you want. You can even draw a picture and argue convincingly from that, as long as you are precise in your statements.

Answer 1

$\small \text{Let } X \text{ and } Y \text{ be the random variables denoting the delay for Alice and Bob respectively. Then, } X \\ \text{ and } Y \text{are jointly distributed according to the pdf}, \\ f(x, y) = \begin{cases} 1; & (x, y) \in [0,1]\times[0,1] \\ 0; & \text{otherwise} \end{cases}\\ \text{That is, the density is uniformly 1 over the unit square } [0,1]^2 \text{ and 0 otherwise}.\\ \\ \text{Now, it is given that the person to arrive will wait for 20 minutes(1/3 of an hour) for the other person}\\ \text{after which they will become impatient and will leave if the other person has not arrived by then. We}\\ \text{want to calculate the probability that they will meet.}\\ \\$

Suppose that Alice comes first and is late by x hours (0<x<1). Then, according to the question, Alice and Bob will be able to meet if and only if the delay for Bob is less than equal to x + 1/3. That is, if and only if, y <x+1/3 where y is the amount of time by which Bob is late. Equivalently this condition can be written as, y - 2 <1/3. On the other hand, now suppose that Bob comes first and is late by y hours. Then as for the above case, he will be able to meet Alice if and only if the delay time r of Alice satisfies the condition, 2-y<1/3 From this discussion, we can conclude that Alice and Bob will be able to meet if and only if the absolute difference between their delay time is less than 1/3 irrespective of who arrives first. And so, P(Alice and Bob meet) = P(X -Y <1/3)
$\small \text{To calculate this probability, follow the discussion in the image below. The final answer is } \\ P(\text{alice and Bob meet}) = 7/9 = 0.778$

Given that, fbi Guy) {i (22.8) E (0,1)*(0,1] otherwise. We want P(1x-41€ 13) het A 1(x,y): 12-y15.13, 0435, osysi} Then, since the density is uniform, P (A) is simply the ana of the ana of the region A. x-y -V3 (2/3,1) Ix-y = 1/3 1 BI (1,213) (0,7) B2 (V3,0) 1 The shaded portion in the above figur sepresentan the region A. and the xguind probability is the ana of this region A. It can be calculated by subtracting the ARA) ana of triangles B & B2 from the total ara. Thus, ArCA) 1 - Ar(B) - Ax (B2) Further, An (B) As (B2). ₂ x - x L x 2 333 ) 3 Ya -> An (A) = - 219 = 7/9 = P(Alice & B of meet) PP (A) = A (A) - 7/9 F0:778 Scanned with CS CamScanner